How do you solve $100 x^{2} = 121$ $100x^2 = 121$ ?

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How do you solve $100 x^{2} = 121$ $100x^2 = 121$ ?

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Answer 1 · 2016-07-21T20:57:49

$x = + \frac{11}{10} or x = - \frac{11}{10}$ $x = +11/10 or x = -11/10$

Explanation:

The general rule for solving a quadratic equation $(x^{2})$ $(x^2")$ is to make it equal to 0.
However this case is the exception because there is no term in $x$ $x$ .

$100 x^{2} = 121 make x^{2} the subject$ $100x^2 = 121" make "x^2" the subject"$

$x^{2} = \frac{121}{100} find the square root$ $x^2 = 121/100" find the square root"$

$x = \pm \sqrt{\frac{121}{100}}$ $x = +-sqrt(121/100)$

$x = + \frac{11}{10} or x = - \frac{11}{10}$ $x = +11/10 or x = -11/10$

We can also follow the usual method to get:

$100 x^{2} - 121 = 0 factorise$ $100x^2 - 121 = 0" factorise"$

$(10 x + 11) (10 x - 11) = 0$ $(10x+11)(10x-11)=0$

$x = - \frac{11}{1} or x = \frac{11}{10}$ $x = -11/1 or x = 11/10$

Answer 2 · 2016-07-21T20:58:56

$x = 1.1$ $x=1.1$
$x = - 1.1$ $x=-1.1$

Explanation:

$100 x^{2} = 121$ $100x^2=121$
OR
$x^{2} = \frac{121}{100}$ $x^2=121/100$
OR
$x = \sqrt{\frac{121}{100}}$ $x=sqrt(121/100)$
OR
$x = \frac{11}{10}$ $x= 11/10$
$x = 1.1$ $x=1.1$
OR
$x = - \frac{11}{10}$ $x=-11/10$
$x = - 1.1$ $x=-1.1$

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How do you solve $100 x^{2} = 121$ $100x^2 = 121$ ?

2 Answers

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